1
$\begingroup$

Let $A=\{1-1/n:n\in\mathbb{N}\}⊂\mathbb{Q}$

I want to proof the supremum and infimum of the set $A$ is. I can intuitively see that it is $sup(A)=1$ and $inf(A)=0$.

I know I need to proof that 1 is in the upper bound and 0 is in the lower bound, and then that it is the smallest upper and largest lower bound.

The upper bound of $A$ is $1$, as $1/n$ tends to 0 when $n$ gets larger. And the lower bound is $0$ for the same reason.

But how do I reason/proof that they are the smallest upper and largest lower bound?

$\endgroup$
  • $\begingroup$ Suppose that there is a smaller upper bound (resp. larger lower bound) and arrive at a contradiction. $\endgroup$ – wjm Oct 2 '17 at 19:04
  • $\begingroup$ Can you use the result that $a$ is the supremum of $A$ if and only if $a$ is an upper bound for $A$ and for all $\varepsilon>0$ we have $(a-\varepsilon,a]\cap A\neq\emptyset$? (similar result for $\inf(A)$). $\endgroup$ – Dave Oct 2 '17 at 19:23
1
$\begingroup$

You want to show that for all $x<1$, $x$ is not an upper bound of $A$, which means there exists some $a\in A$ such that $a>x$. Finding this $a$ will have to depend on $x$.

Similarly, you want to show that for all $x>0$, $x$ is not a lower bound of $A$, meaning there exists some $a\in A$ such that $a<x$. Finding this $a$ will be easy.

$\endgroup$
  • $\begingroup$ So if I let $a=1$ I have $1>x$. Is writing this (in a more proper way) a correct way of showing that $sup(A)=1$? $\endgroup$ – Sirmimer Oct 2 '17 at 19:28
  • $\begingroup$ @Sirmimer Well, you can't take $a=1$, since $1\not\in A$. $\endgroup$ – Chris Culter Oct 2 '17 at 19:32
  • $\begingroup$ I thought $A=[0,1]$ as $1/n$ approaches 0 and therefore you have $1-0=1$? $\endgroup$ – Sirmimer Oct 2 '17 at 19:36
  • $\begingroup$ @Sirmimer $1/n$ approaches $0$, but there is no value of $n$ for which it equals $0$. So $1$ is not a member of the set $A$. $\endgroup$ – Chris Culter Oct 2 '17 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.